Quantile-based random sparse Kaczmarz for corrupted and noisy linear systems

Abstract

The randomized Kaczmarz method, along with its recently developed variants, has become a popular tool for dealing with large-scale linear systems. However, these methods usually fail to converge when the linear systems are affected by heavy corruption, which is common in many practical applications. In this study, we develop a new variant of the randomized sparse Kaczmarz method with linear convergence guarantees, by making use of the quantile technique to detect corruptions. Moreover, we incorporate the averaged block technique into the proposed method to achieve parallel computation and acceleration. Finally, the proposed algorithms are illustrated to be very efficient through extensive numerical experiments.

Appendix

Proof of Lemma 5

Denote the set of all the indices of corrupted equations as C. The set excluding the $\beta m$ corrupted rows is denoted as E. For all $i\in E$, we have $b_i=\tilde{b}_i+r_i$ and $\langle a_i,\hat{x}\rangle =\tilde{b}_i=b_i-r_i$. Hence,

$$ |\langle a_i,x_k-\hat{x}\rangle | =|\langle a_i,x_k\rangle -b_i+r_i|\nonumber \ge |\langle a_i,x_k\rangle -b_i|-\Vert r\Vert _{\infty }. $$

Thus,

$$\begin{aligned} |\langle a_i,x_k\rangle -b_i|\le |\langle a_i,x_k-\hat{x}\rangle |+\Vert r\Vert _{\infty }. \end{aligned}$$

(1)

Recall that $Q_k=Q_q(x_{k},1\le i\le m )$ and $|E|=(1-\beta )m$, meaning that at least $(1-\beta -q)m$ numbers of residuals $|\langle a_i,x_k\rangle -b_i|_{i=1}^m$ are at least $Q_k$ and have not been corrupted. Then,

$$\begin{aligned} (1-\beta -q)mQ_k&\le \sum _{i\in E}|\langle a_i,x_k\rangle -b_i|\nonumber \\&\le \sum _{i\in E}(|\langle a_i,x_k-\hat{x}\rangle |+\Vert r\Vert _{\infty })\nonumber \\&\le \left( \sum _{i\in E}|\langle a_i,x_k-\hat{x}\rangle |^2\right) ^{\frac{1}{2}}\sqrt{|E|}+|E|\cdot \Vert r\Vert _{\infty }\nonumber \\&= \Vert A_{ E}(x_k-\hat{x})\Vert \sqrt{(1-\beta )m} +(1-\beta )m\Vert r\Vert _{\infty }\nonumber \\&\le \sigma _{\max }\Vert x_k-\hat{x}\Vert \sqrt{(1-\beta )m}+ (1-\beta )m\Vert r\Vert _{\infty }. \end{aligned}$$

(2)

Therefore,

$$\begin{aligned} Q_k\le \frac{\sqrt{1-\beta }}{(1-\beta -q)\sqrt{m}}\sigma _{\max }\Vert x_k-\hat{x}\Vert + \frac{1-\beta }{1-\beta -q}\Vert r\Vert _{\infty }, \end{aligned}$$

(3)

which completes the proof. $\square$

Proof of Theorem 1

We start by introducing some useful notations. Denote the set consisting of all acceptable indices in the k-th iterate as

$$ B=\{1\le i\le m\mid |\langle x_k,a_i\rangle | \le Q_k\}.\nonumber $$

The subset consisting of corrupted indices in B is denoted as S; then the other indices in B are in subset $B\backslash S$. Apparently, $(q-\beta )m\le |B\backslash S|\le qm$ and $0\le |S|\le \beta m$. Note that the used index in each iterate is sampled from the acceptable set B in each iterate. This observation inspires us to consider the following splitting:

$$\begin{aligned} \mathbb {E}\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\right] =&P(i\in S)\mathbb {E}\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})|i\in S\right] \nonumber \\&+ P(i\in B\backslash S)\mathbb {E}\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})|i\in B\backslash S\right] . \end{aligned}$$

(4)

Our remainder assignment is to estimate the split terms above. For clarity, we respectively take both inexact and exact steps into account, which are all divided into three steps.

(a) In the inexact-step case, firstly we consider the uncorrupted equations indexed $i\in B\backslash S$. Since $b_{B\backslash S}^c=0$, the equations indexed $i\in B\backslash S$ satisfy

$$ A_{B\backslash S}x=\tilde{b}_{B\backslash S}=b_{B\backslash S}-r_{B\backslash S}. $$

According to the convergence rate of RaSK in noisy case in Lemma 4(a), we have

$$\begin{aligned}&~~~~~\mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})|i\in B\backslash S\right] \nonumber \\&\le \left( 1-\frac{1}{2}\cdot \frac{1}{\tilde{\kappa }(A_{B\backslash S})^2}\cdot \frac{|\hat{x}|_{\text {min}}}{|\hat{x}|_{\text {min}}+2\lambda }\right) D_f^{x_{k}^*}(x_{k},\hat{x})+\frac{1}{2}\frac{\Vert r_{B\backslash S}\Vert ^2}{\Vert A_{B\backslash S}\Vert _F^2}\nonumber \\&\le \left( 1-\frac{1}{2}\cdot \frac{\tilde{\sigma }_{q-\beta ,\text {min}}^2}{qm}\cdot \frac{|\hat{x}|_{\text {min}}}{|\hat{x}|_{\text {min}}+2\lambda }\right) D_f^{x_{k}^*}(x_{k},\hat{x}) +\frac{1}{2}\Vert r\Vert _{\infty }^2, \end{aligned}$$

(5)

where the last inequality follows from $\Vert r_{B\backslash S}\Vert ^2\le |B\backslash S|\cdot \Vert r\Vert _{\infty }^2$ and

$$ \tilde{\kappa }(A_{B\backslash S})^2= \frac{\Vert A_{B\backslash S}\Vert _F^2}{\tilde{\sigma }^2_{\text {min}}(A_{B\backslash S})}\le \frac{qm}{\tilde{\sigma }_{q-\beta ,\text {min}}^2}. $$

Second, we consider the conditional expectation in S. In this case,

$$ A_Sx=\tilde{b}_S, b_S=\tilde{b}_S+b_S^c+r_S. $$

Denote the orthogonal projection of true solution $\hat{x}\in H(a_{i_k},\tilde{b}_{i_k})$ onto corrupted hyperplane $H(a_{i_k},b_{i_k})$ as $x_k^c$ [9], implying that

$$\begin{aligned} x_k^c:=\hat{x}+(b_{i_k}-\tilde{b}_{i_k})a_{i_k}\in H(a_{i_k},b_{i_k}). \end{aligned}$$

(6)

Note that $x_{k+1}$ is the Bregman projection of $x_k$ onto the hyperplane $H(a_{i_k},b_{i_k})$. According to Lemma 2 and the 1-strong convexity of f, it follows that

$$\begin{aligned} D_f^{x_{k+1}^*}(x_{k+1},x_k^c) \le D_f^{x_{k}^*}(x_{k},x_k^c) - \frac{1}{2}(\langle a_{i_k},x_k\rangle -b_{i_k})^2. \end{aligned}$$

(7)

By reformulating it, we obtain that

$$\begin{aligned} D_f^{x_{k+1}^*}(x_{k+1},\hat{x}) \le D_f^{x_{k}^*}(x_{k},\hat{x}) - \frac{1}{2}(\langle a_{i_k},x_k\rangle -b_{i_k})^2 +\langle x_{k+1}^*-x_{k}^*,x_k^c-\hat{x}\rangle . \end{aligned}$$

(8)

For Quantile-RaSK with inexact step, we have $x_{k+1}^*-x_{k}^*=-(\langle a_{i_k},x_k\rangle -b_{i_k})a_{i_k}$, obtained from the 11-step of Algorithm 1. Hence, (8) can be rewritten as

$$\begin{aligned}&~~~~D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\nonumber \\&\le D_f^{x_{k}^*}(x_{k},\hat{x}) + \frac{1}{2}(\langle a_{i_k},x_k\rangle -b_{i_k})^2 -(\langle a_{i_k},x_k\rangle -b_{i_k})(\langle a_{i_k},x_k\rangle -\tilde{b}_{i_k}). \end{aligned}$$

(9)

Now fix the values of the indices $i_0,\ldots ,i_{k-1}$ and consider only $i_k$ as a random variable with values in $\{1,\ldots ,m\}$. According to $|\langle a_{i_k},x_k\rangle -b_{i_k}|\le Q_k$, we have

$$\begin{aligned}&~~~~\mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})|i\in S\right] \!\le \! D_f^{x_{k}^*}(x_{k},\hat{x}) \!+\!\frac{1}{2}Q_k^2\!+\!Q_k\cdot \mathbb {E}_k\left( |\langle a_i,x_k\!-\!\hat{x}\rangle |\mid i\in \! S\right) , \end{aligned}$$

(10)

where

$$\begin{aligned} \mathbb {E}_k\left( |\langle a_i,x_k-\hat{x}\rangle |\mid i\in S\right) \!=\frac{1}{|S|}\sum _{i\in S} |\langle a_i,x_k-\hat{x}\rangle | \!\le \! \frac{1}{\sqrt{|S|}}\Vert A_Sx_k-b_S\Vert \!\le \!\frac{\sigma _{\max }}{\sqrt{|S|}}\Vert x_k-\hat{x}\Vert . \end{aligned}$$

(11)

Combining (3), (10) and (1), we have that

$$\begin{aligned}&~~~~\mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})|i\in S\right] \nonumber \\&\le D_f^{x_{k}^*}(x_{k},\hat{x}) +\frac{1}{2}\left( \frac{1-\beta }{1-\beta -q}\right) ^2\Vert r\Vert _{\infty }^2\nonumber \\&~~~+\frac{1-\beta }{1-\beta -q}\left( \frac{1}{\sqrt{|S|}\sqrt{1-\beta }\sqrt{m}}+\frac{1}{2}\cdot \frac{1}{(1-\beta -q)m} \right) \sigma _{\max }^2\Vert x_k-\hat{x}\Vert ^2\nonumber \\&~~~+\frac{1-\beta }{1-\beta -q}\left( \frac{\sqrt{1-\beta }}{(1-\beta -q)\sqrt{m}}+\frac{1}{\sqrt{|S|}}\right) \sigma _{\max }\Vert x_k-\hat{x}\Vert \cdot \Vert r\Vert _{\infty } . \end{aligned}$$

(12)

To handle the $\Vert x_k-\hat{x}\Vert \cdot \Vert r\Vert _{\infty }$ term we split into two cases: $\Vert x_k-\hat{x}\Vert \ge \sqrt{n} \Vert r\Vert _{\infty }$ and $\Vert x_k-\hat{x}\Vert \le \sqrt{n} \Vert r\Vert _{\infty }$. It is easy to obtain that

$$\begin{aligned}&~~~~\mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\mid i\in S\right] \nonumber \\&\le \left( 1+\frac{c_{A,\beta ,q}'}{\sqrt{|S|}}+C_{A,\beta ,q}\right) D_f^{x_{k}^*}(x_{k},\hat{x}) +\left( \frac{d_{A,\beta ,q}}{\sqrt{|S|}}+D_{A,\beta ,q}\right) \Vert r\Vert _{\infty }^2, \end{aligned}$$

(13)

where

$$\begin{aligned} c_{A,\beta ,q}'= & {} \frac{2\sigma _{\max }^2\sqrt{1-\beta }}{\sqrt{m}(1-\beta -q)}+\frac{2\sigma _{\max }(1-\beta )}{\sqrt{n}(1-\beta -q)}, C_{A,\beta ,q}= \frac{\sigma _{\max }^2(1-\beta )}{m(1-\beta -q)^2} + \frac{2(1-\beta )^{\frac{3}{2}}\sigma _{\max }}{\sqrt{mn}(1-\beta -q)^2},\\ d_{A,\beta ,q}= & {} \frac{(1-\beta )\sqrt{n}}{1-\beta -q}\sigma _{\max },D_{A,\beta ,q}=\frac{(1-\beta )^{\frac{3}{2}}\sqrt{n}}{(1-\beta -q)^2\sqrt{m}}\sigma _{\max }+\frac{1}{2}\left( \frac{1-\beta }{1-\beta -q}\right) ^2. \end{aligned}$$

Finally, combining (4), (5) and (13) we have

$$\begin{aligned} \mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\right] \le \left( 1-C_1\right) D_f^{x_{k}^*}(x_{k},\hat{x}) +C_2\Vert r\Vert _{\infty }^2, \end{aligned}$$

(14)

where

$$\begin{aligned} C_1= & {} \frac{q-\beta }{2q^2} \frac{|\hat{x}|_{\text {min}}}{|\hat{x}|_{\text {min}}+2\lambda } \frac{\tilde{\sigma }_{q-\beta ,\text {min}}^2}{m} - \frac{2\sqrt{\beta (1-\beta )}}{q(1-\beta -q)}\left( \frac{\sigma _{\max }^2}{m}+\frac{\sqrt{1-\beta }\sigma _{\max }}{\sqrt{mn}}\right) \\{} & {} -\frac{\beta (1-\beta )}{q(1-\beta -q)^2}\left( \frac{\sigma _{\max }^2}{m}+\frac{2\sqrt{1-\beta }\sigma _{\max }}{\sqrt{mn}} \right) ,\\ C_2= & {} \frac{\sqrt{\beta }(1-\beta )}{q(1-q-\beta )}\left( 1+\frac{\sqrt{\beta (1-\beta )}}{1-q-\beta }\right) \sqrt{\frac{n}{m}}\sigma _{\max }+\frac{1}{2}\frac{\beta (1-\beta )^2}{q(1-\beta -q)^2}+\frac{1}{2}. \end{aligned}$$

We consider all indices $i_0,i_1,\ldots ,i_k$ as random variables, and take full expectation on both sides. Thus,

$$\begin{aligned} \mathbb {E} \left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\right] \le \left( 1-C_1\right) \mathbb {E}\left[ D_f^{x_{k}^*}(x_{k},\hat{x})\right] +C_2\Vert r\Vert _{\infty }^2, \end{aligned}$$

(15)

(b) In the exact-step case, first we consider the uncorrupted equations indexed $i\in B\backslash S$. Since $b_{B\backslash S}^c=0$, the equations indexed $i\in B\backslash S$ satisfy

$$ A_{B\backslash S}x=\tilde{b}_{B\backslash S}=b_{B\backslash S}-r_{B\backslash S}. $$

It follows from the convergence rate of ERaSK for noisy case in Lemma 4(b) that

$$\begin{aligned}&~~~~\mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})|i\in B\backslash S\right] \nonumber \\&\le \left( 1-\frac{1}{2}\frac{\tilde{\sigma }_{q-\beta ,\text {min}}^2}{qm}\frac{|\hat{x}|_{\text {min}}}{|\hat{x}|_{\text {min}}+2\lambda }\right) D_f^{x_{k}^*}(x_{k},\hat{x}) +\frac{1}{2}\Vert r\Vert _{\infty }^2 +\frac{2}{\sqrt{|B\backslash S|}}\Vert r\Vert _{\infty }\Vert A\Vert _{1,2}. \end{aligned}$$

(16)

Second, we consider the conditional expectation in corrupted set S. In this case, we have

$$ A_Sx=\tilde{b}_S, b_S=\tilde{b}_S+b_S^c+r_S.\nonumber $$

For Quantile-RaSK with exact step, we have $x_{k}^*=x_{k}+\lambda s_k$ with $\Vert s_k\Vert _{\infty }\le 1$ and $\Vert s_{k+1}\Vert _{\infty }\le 1$; then $x_{k+1}^*-x_{k}^*=(x_{k+1}-x_{k})+\lambda (s_{k+1}-s_{k})$. Note that the exact linesearch guarantees $\langle a_{i_k},x_{k+1}\rangle =b_{i_k}$. Thus, (8) can be rewritten as

$$\begin{aligned} D_f^{x_{k+1}^*}(x_{k+1},\hat{x})&\le D_f^{x_{k}^*}(x_{k},\hat{x}) + \lambda \langle s_{k+1}-s_{k},a_{i_k}\rangle (b_{i_k}-\tilde{b}_{i_k})\nonumber \\&~~~+ \frac{1}{2}(\langle a_{i_k},x_k\rangle -b_{i_k})^2 - (\langle a_{i_k},x_k\rangle -b_{i_k}) (\langle a_{i_k},x_k\rangle -\tilde{b}_{i_k}). \end{aligned}$$

(17)

Viewing $i_k$ as a random variable with fixed $i_0,\ldots ,i_{k-1}$, yields

$$\begin{aligned} \mathbb {E}_k(\lambda \langle s_{k+1}-s_{k},a_i\rangle (b_i-\tilde{b}_i)|i\in S)&\le 2\lambda \mathbb {E}_k(\Vert a_i\Vert _1\cdot |b_i-\tilde{b}_i||i\in S)\nonumber \\&\le \frac{2\lambda }{|S|}\sum _{i\in S}(\Vert a_i\Vert _1\cdot |b_i-\tilde{b}_i|)\nonumber \\&\le \frac{2\lambda }{|S|}\Vert b-\tilde{b}\Vert \cdot \Vert A\Vert _{1,2}. \end{aligned}$$

(18)

And recall the conclusion (13) in (a), we obtain

$$\begin{aligned}&~~~\mathbb {E}_k\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\mid i\in S\right] \nonumber \\&\le \left( 1+\frac{c_{A,\beta ,q}'}{\sqrt{|S|}}+C_{A,\beta ,q}\right) D_f^{x_{k}^*}(x_{k},\hat{x}) +\left( \frac{d_{A,\beta ,q}}{\sqrt{|S|}}+D_{A,\beta ,q}\right) \Vert r\Vert _{\infty }^2\nonumber \\&~~~ +\frac{2\lambda }{|S|}\Vert b-\tilde{b}\Vert \cdot \Vert A\Vert _{1,2}. \end{aligned}$$

(19)

Finally, combining all ingredients: (4), (16) and (19), and taking full expectation, we have that

$$\begin{aligned}&~~~~\mathbb {E}\left[ D_f^{x_{k+1}^*}(x_{k+1},\hat{x})\right] \\&\le \left( 1-C_1\right) \mathbb {E}\left[ D_f^{x_{k}^*}(x_{k},\hat{x})\right] +C_2\Vert r\Vert _{\infty }^2 +\frac{2}{\sqrt{qm}}\Vert r\Vert _{\infty }\cdot \Vert A\Vert _{1,2} +\frac{2\lambda }{qm}\Vert b-\tilde{b}\Vert \cdot \Vert A\Vert _{1,2}. \end{aligned}$$

The constants $C_1$ and $C_2$ are placed in Theorem 1.

(c) To ensure the decay in expectation, we require

$$ \left( \frac{2\sqrt{(1-\beta )\beta }}{1-\beta -q}+\frac{(1-\beta )\beta }{(1-\beta -q)^2} \right) \cdot \sigma _{\max } + \sqrt{\frac{m}{n}}\left( \frac{2\sqrt{\beta }(1-\beta )}{1-\beta -q} + \frac{2(1-\beta )^{\frac{3}{2}}\beta }{(1-\beta -q)^2} \right) \nonumber $$

$$ < \frac{q-\beta }{2q}\cdot \frac{|\hat{x}|_{\text {min}}}{|\hat{x}|_{\text {min}}+2\lambda }\cdot \frac{\tilde{\sigma }_{q-\beta ,\text {min}}^2}{\sigma _{\max }},\nonumber $$

which holds for a small enough parameter $\beta$ since the left-hand side of it tends to zero as $\beta$ tends to zero. Therefore, we obtain the conclusion. $\square$

The proof of Theorem 3

Similar to the proof of Theorem 1, we first denote the set of row indices less than the quantile $Q_k$ at iteration k as $T=\{i\in [m]\mid |\langle a_i,x_k\rangle -b_i|<Q_k\},|T|=\eta =qm$. The uncorrupted and corrupted rows in T are respectively denoted as $T_1$ and $T_2$. Then $(q-\beta )m\le |T_1|\le qm,|T_2|\le \beta m$. Using the constant stepsize, the update in Algorithm 2 is as follows:

$$\begin{aligned} x_{k+1}^*= & {} x_k^*-\frac{w}{\eta }\sum _{i\in T} (\langle a_i,x_k\rangle -b_i) a_{i},\nonumber \\ x_{k+1}= & {} \mathcal {S}_{\lambda }(x_{k+1}^*). \end{aligned}$$

(20)

Denote

$$\begin{aligned} x_{k}^{\delta }:=\hat{x}+\frac{w}{\eta }\sum _{i\in T} (b_i-\tilde{b}_i) a_{i}. \end{aligned}$$

(21)

Now use Lemma 8 with $f(x)=\lambda \Vert x\Vert _{1}+\frac{1}{2}\Vert x\Vert ^2$, $\Phi (x)=\langle x_k^*-x_{k+1}^*,x-x_k\rangle$ and $y=x_k^{\delta }$, and it holds that

$$\begin{aligned}{} & {} D_f^{x_{k+1}^*}(x_{k+1}, x_k^{\delta })\\\le & {} D_f^{x_k^*}(x_k, x_k^{\delta }) \!+\langle x_k^{*}\!-x_{k+1}^{*},x_k^{\delta }\!-x_k\rangle \!-\langle x_k^*\!-x_{k+1}^*,x_{k+1}\!-x_k\rangle \!-D_f^{x_{k+1}^*}(x_{k},x_{k+1})\\\le & {} D_f^{x_k^*}(x_k, x_k^{\delta }) \!+\!\langle x_k^*\!-\!x_{k+1}^*,x_k^{\delta }\!-\!x_k\rangle \!+\!\Vert x_k^*\!-\!x_{k+1}^*\Vert \cdot \Vert x_{k+1}\!-\!x_k\Vert \!-\!\frac{1}{2} \Vert x_k\!-\!x_{k+1}\Vert ^2\\\le & {} D_f^{x_k^*}(x_k, x_k^{\delta }) +\langle x_k^*-x_{k+1}^*,x_k^{\delta }-x_k\rangle +\frac{1}{2} \Vert x_k^*-x_{k+1}^*\Vert ^2, \end{aligned}$$

Unfolding the expression of $x_k^{\delta }$ in (21), we obtain

$$\begin{aligned} \begin{aligned}&~~~~D_f^{x_{k+1}^*}\left( x_{k+1}, \hat{x}\right) \\&\le D_f^{x_k^*}\left( x_k, \hat{x}\right) +\langle x_{k}^*-x_{k+1}^*,\hat{x}-x_k\rangle +\frac{1}{2} \Vert x_k^*-x_{k+1}^*\Vert ^2, \end{aligned} \end{aligned}$$

We divide it into two steps: the scalar product and the quadratic term.

Step 1: For the inner product term, we can derive that

$$\begin{aligned} \begin{aligned}&~~~~\langle x_{k}^*-x_{k+1}^*,\hat{x}-x_k\rangle \\&= -\langle \frac{w}{\eta }\sum _{i\in T} (\langle a_i,x_k\rangle -b_i) a_{i},x_k-\hat{x}\rangle \\&= -\langle \frac{w}{\eta }\sum _{i\in T_1} (\langle a_i,x_k\rangle -b_i) a_{i},x_k-\hat{x}\rangle -\langle \frac{w}{\eta }\sum _{i\in T_2} (\langle a_i,x_k\rangle -b_i) a_{i},x_k-\hat{x}\rangle \\ \end{aligned} \end{aligned}$$

(22)

From the first term in (22), we obtain

$$\begin{aligned}&~~~~-\langle \frac{w}{\eta }\sum _{i\in T_1}(\langle a_i,x_k\rangle -b_i) a_{i},x_k-\hat{x}\rangle \nonumber \\&= -\langle \frac{w}{\eta }\sum _{i\in T_1} (\langle a_i,x_k\rangle -\tilde{b}_i) a_{i},x_k-\hat{x}\rangle +\langle \frac{w}{\eta }\sum _{i\in T_1} r_i a_{i},x_k-\hat{x}\rangle \nonumber \\&\le -\frac{w}{\eta }\langle x_k-\hat{x},\sum _{i\in T_1}a_ia_i^T(x_k-\hat{x})\rangle +\frac{w}{\eta }\Vert r\Vert _{\infty }\langle \sum _{i\in T_1} a_{i},x_k-\hat{x}\rangle \nonumber \\&\le -\frac{w}{qm}\sigma _{q-\beta ,\text {min}}^2\cdot \Vert x_k-\hat{x}\Vert ^2 + \frac{w}{\sqrt{qm}}\sigma _{\max }\cdot \Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert . \end{aligned}$$

(23)

From the second term in (22), we get

$$\begin{aligned}&~~~~-\langle \frac{w}{\eta }\sum _{i\in T_2} (\langle a_i,x_k\rangle -b_i) a_{i},x_k-\hat{x}\rangle \nonumber \\&\le \frac{w}{\eta }Q_k\langle \sum _{i\in T_2} a_{i},x_k-\hat{x}\rangle \nonumber \\&\le \frac{w}{\eta }Q_k\sqrt{| T_2|}\sigma _{\max }\cdot \Vert x_k-\hat{x}\Vert \nonumber \\&\le \frac{w\sqrt{(1-\beta )\beta }}{(1-\beta -q)qm}\sigma _{\max }^2\cdot \Vert x_k-\hat{x}\Vert ^2 + \frac{1-\beta }{1-\beta -q} \frac{w\sqrt{\beta }}{q\sqrt{m}}\sigma _{\max }\cdot \Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert . \end{aligned}$$

(24)

The last inequality makes use of Lemma 5. Combining (22), (23) and (24), we obtain

$$\begin{aligned}&~~~~\langle x_{k}^*-x_{k+1}^*,\hat{x}-x_k\rangle \nonumber \\&\le \left( -\frac{w}{qm}\sigma _{q-\beta ,\text {min}}^2 +\frac{w\sqrt{(1-\beta )\beta }}{(1-\beta -q)qm}\sigma _{\max }^2 \right) \Vert x_k-\hat{x}\Vert ^2\nonumber \\&~~~~+ \left( \frac{w}{\sqrt{qm}}\sigma _{\max } + \frac{1-\beta }{1-\beta -q} \frac{w\sqrt{\beta }}{q\sqrt{m}}\sigma _{\max } \right) \Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert . \end{aligned}$$

(25)

Step 2: For the quadratic term, we have

$$\begin{aligned}&~~~~\Vert x_k^*-x_{k+1}^*\Vert ^2\nonumber \\&=\Vert \frac{w}{\eta }\sum _{i\in T} (\langle a_i,x_k\rangle -b_i) a_{i}\Vert ^2\nonumber \\&=\Vert \frac{w}{\eta }\sum _{i\in T_1} (\langle a_i,x_k\rangle -b_i) a_{i}+\frac{w}{\eta }\sum _{i\in T_2} (\langle a_i,x_k\rangle -b_i) a_{i}\Vert ^2,\nonumber \\&=\Vert u+v\Vert ^2,\nonumber \\&\le (\Vert u\Vert +\Vert v\Vert )^2, \end{aligned}$$

(26)

where $u=\frac{w}{\eta }\sum _{i\in T_1} (\langle a_i,x_k\rangle -b_i) a_{i},v=\frac{w}{\eta }\sum _{i\in T_2} (\langle a_i,x_k\rangle -b_i) a_{i}$.

We have

$$\begin{aligned} \Vert u\Vert ^2&=\Vert \frac{w}{\eta }\sum _{i\in T_1} (\langle a_i,x_k\rangle -b_i) a_{i}\Vert ^2\nonumber \\&= \frac{w^2}{\eta ^2}\Vert \sum _{i\in T_1} (\langle a_i,x_k\rangle -\tilde{b}_i-r_i) a_{i}\Vert ^2\nonumber \\&= \frac{w^2}{\eta ^2} \Vert A_{ T_1}^T A_{ T_1}(x_k-\hat{x})-\sum _{i\in T_1}a_ir_i \Vert ^2\nonumber \\&\le \frac{w^2}{\eta ^2} \left( \Vert A_{ T_1}^T A_{ T_1}(x_k-\hat{x})\Vert +\Vert \sum _{i\in T_1}a_ir_i\Vert \right) ^2\nonumber \\&\le \frac{w^2\sigma _{\max }^2}{q^2m^2}\left( \sigma _{\max }\Vert x_k-\hat{x}\Vert +\sqrt{qm}\Vert r\Vert _{\infty } \right) ^2. \end{aligned}$$

(27)

and

$$\begin{aligned} \Vert v\Vert ^2&=\Vert \frac{w}{\eta }\sum _{i\in T_2} (\langle a_i,x_k\rangle -b_i) a_{i}\Vert ^2\nonumber \\&\le \frac{w^2}{\eta ^2}\cdot Q_k^2\cdot \Vert \sum _{i\in T_2}a_i\Vert ^2\nonumber \\&\le \frac{w^2\sigma _{\max }^2\beta }{q^2m}\left( \frac{\sqrt{1-\beta }}{(1-\beta -q)\sqrt{m}}\sigma _{\max }\Vert x_k-\hat{x}\Vert + \frac{1-\beta }{1-\beta -q}\Vert r\Vert _{\infty } \right) ^2. \end{aligned}$$

(28)

Bring (27) and (28) together, we obtain

$$\begin{aligned}&~~~~\Vert u+v\Vert ^2\nonumber \\&\le \frac{w^2\sigma _{\max }^2}{q^2m^2}\left( \sigma _{\max }\Vert x_k-\hat{x}\Vert +\sqrt{qm}\Vert r\Vert _{\infty } + \frac{\sqrt{(1-\beta )\beta }}{1-\beta -q}\sigma _{\max }\Vert x_k-\hat{x}\Vert + \frac{(1-\beta )\sqrt{\beta m}}{1-\beta -q}\Vert r\Vert _{\infty } \right) ^2\nonumber \\&= \frac{w^2\sigma _{\max }^2}{q^2m^2} \left[ \sigma _{\max } \left( 1+\frac{\sqrt{(1-\beta )\beta }}{1-\beta -q} \right) \Vert x_k-\hat{x}\Vert + \sqrt{qm} \left( 1+\frac{(1-\beta )\sqrt{\beta }}{(1-\beta -q)\sqrt{q}} \right) \Vert r\Vert _{\infty } \right] ^2\nonumber \\&\le \frac{w^2}{q^2m^2}\sigma _{\max }^4\left( 1+ \frac{\sqrt{(1-\beta )\beta }}{1-\beta -q} \right) ^2\Vert x_k-\hat{x}\Vert ^2 + \frac{w^2}{qm}\sigma _{\max }^2\left( 1+\frac{(1-\beta )\sqrt{\beta }}{(1-\beta -q)\sqrt{q}} \right) ^2\Vert r\Vert _{\infty }^2\nonumber \\&~~~~+ \frac{2w^2}{(qm)^{\frac{3}{2}}}\sigma _{\max }^3 \left( 1+ \frac{\sqrt{(1-\beta )\beta }}{1-\beta -q} \right) \left( 1+\frac{(1-\beta )\sqrt{\beta }}{(1-\beta -q)\sqrt{q}} \right) \Vert x_k-\hat{x}\Vert \cdot \Vert r\Vert _{\infty }. \end{aligned}$$

(29)

Combining (25) and (29) yields

$$\begin{aligned}{} & {} D_f^{x_{k+1}^*}\left( x_{k+1}, \hat{x}\right) \\\le & {} D_f^{x_k^*}\left( x_k, \hat{x}\right) +\langle x_{k}^*-x_{k+1}^*,\hat{x}-x_k\rangle +\frac{1}{2} \Vert x_k^*-x_{k+1}^*\Vert ^2,\nonumber \\\le & {} D_f^{x_k^*}\left( x_k, \hat{x}\right) + \left( -\frac{w}{qm}\sigma _{q-\beta ,\text {min}}^2 +\frac{w\sqrt{(1-\beta )\beta }}{(1-\beta -q)qm}\sigma _{\max }^2 \right) \Vert x_k-\hat{x}\Vert ^2\nonumber \\{} & {} + \left( \frac{w}{\sqrt{qm}}\sigma _{\max } + \frac{1-\beta }{1-\beta -q} \frac{w\sqrt{\beta }}{q\sqrt{m}}\sigma _{\max } \right) \Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert \nonumber \\{} & {} + \frac{w^2}{q^2m^2}\sigma _{\max }^4\left( 1+ \frac{\sqrt{(1-\beta )\beta }}{1-\beta -q} \right) ^2\Vert x_k-\hat{x}\Vert ^2 + \frac{w^2}{qm}\sigma _{\max }^2\left( 1+\frac{(1-\beta )\sqrt{\beta }}{(1-\beta -q)\sqrt{q}} \right) ^2\Vert r\Vert _{\infty }^2\nonumber \\{} & {} + \frac{2w^2}{(qm)^{\frac{3}{2}}}\sigma _{\max }^3 \left( 1+ \frac{\sqrt{(1-\beta )\beta }}{1-\beta -q} \right) \left( 1+\frac{(1-\beta )\sqrt{\beta }}{(1-\beta -q)\sqrt{q}} \right) \Vert x_k-\hat{x}\Vert \cdot \Vert r\Vert _{\infty }\\= & {} D_f^{x_k^*}\left( x_k, \hat{x}\right) + \left( -c_1\frac{w}{m}\sigma _{q-\beta ,\text {min}}^2 +c_2\frac{ w}{m}\sigma _{\max }^2 +c_3 \frac{w^2}{m^2}\sigma _{\max }^4 \right) \Vert x_k-\hat{x}\Vert ^2\nonumber \\{} & {} + \left( c_4\frac{w}{\sqrt{m}}\sigma _{\max } +c_5\frac{w^2}{m^{\frac{3}{2}}}\sigma _{\max }^3 \right) \Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert + c_6\frac{w^2}{m}\sigma _{\max }^2\Vert r\Vert _{\infty }^2, \end{aligned}$$

where $c_i,i=1,\ldots ,6$ are positive constants only related to q and $\beta$; they can be directly obtained and we omit it. For the term $\Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert$, we use average inequality

$$\Vert r\Vert _{\infty }\cdot \Vert x_k-\hat{x}\Vert \le \frac{1}{2\sqrt{n}}\Vert x_k-\hat{x}\Vert ^2+\frac{\sqrt{n}}{2}\Vert r\Vert _{\infty }^2.$$

Denote $\alpha =\frac{|\hat{x}|_{\text {min}}}{ |\hat{x}|_{\text {min}} +2 \lambda }$, and use Lemma 9, we have

$$\begin{aligned}&~~~~D_f^{x_{k+1}^*}\left( x_{k+1}, \hat{x}\right) \nonumber \\&\le D_f^{x_k^*}\left( x_k, \hat{x}\right) + \left( -c_1\frac{w}{m}\sigma _{q-\beta ,\text {min}}^2 +c_2\frac{w}{m}\sigma _{\max }^2 +c_3\frac{w^2}{m^2}\sigma _{\max }^4 +\frac{c_4}{2}\frac{w}{\sqrt{mn}}\sigma _{\max } \right) \Vert x_k-\hat{x}\Vert ^2\nonumber \\&~~~~~~ +\frac{c_5}{2}\frac{w^2}{m^{3/2}n^{1/2}}\sigma _{\max }^3\Vert x_k-\hat{x}\Vert ^2 + \left( \frac{c_4}{2}\frac{\sqrt{n}w}{\sqrt{m}}\sigma _{\max }+\frac{c_5}{2}\frac{\sqrt{n}w^2}{m^{3/2}}\sigma _{\max }^3+c_6\frac{w^2}{m}\sigma _{\max }^2 \right) \Vert r\Vert _{\infty }^2 \nonumber \\&\le \left[ 1-\alpha \left( c_1\frac{w}{m}\frac{\tilde{\sigma }_{\text {min}}^2\sigma _{q-\beta ,\text {min}}^2}{\sigma _{\max }^2} -c_2\frac{w}{m}\tilde{\sigma }_{\text {min}}^2 -c_3\frac{w^2}{m^2}\tilde{\sigma }_{\text {min}}^2\sigma _{\max }^2 -\frac{c_4}{2}\frac{w\tilde{\sigma }_{\text {min}}^2}{\sqrt{mn}\sigma _{\max }} \right) \right] D_f^{x_k^*}\left( x_k, \hat{x}\right) \nonumber \\&~~~~~~ +\frac{c_5}{2}\frac{\alpha w^2\tilde{\sigma }_{\text {min}}^2}{m^{3/2}n^{1/2}}\sigma _{\max }D_f^{x_k^*}\left( x_k, \hat{x}\right) + \left( \frac{c_4}{2}\frac{\sqrt{n}w}{\sqrt{m}}\sigma _{\max }+\frac{c_5}{2}\frac{\sqrt{n}w^2}{m^{3/2}}\sigma _{\max }^3+c_6\frac{w^2}{m}\sigma _{\max }^2 \right) \Vert r\Vert _{\infty }^2\nonumber \\&=(1-c_1^*w+c_2^*w^2)D_f^{x_k^*}\left( x_k, \hat{x}\right) +(c_3^*w+c_4^*w^2)\Vert r\Vert _{\infty }^2, \end{aligned}$$

(30)

where

$$\begin{aligned} c_1^*= & {} c_1\alpha \frac{\tilde{\sigma }_{\text {min}}^2\sigma _{q-\beta ,\text {min}}^2}{ m\sigma _{\max }^2}-c_2\alpha \frac{\tilde{\sigma }_{\text {min}}^2}{ m}-c_4\alpha \frac{\tilde{\sigma }_{\text {min}}^2}{2\sqrt{mn}\sigma _{\max }},~ c_2^*=c_3\alpha \frac{\tilde{\sigma }_{\text {min}}^2\sigma _{\max }^2}{ m^2}+c_5\alpha \frac{\tilde{\sigma }_{\text {min}}^2\sigma _{\max }}{2 m^{3/2}n^{1/2}},\\ c_3^*= & {} c_4\frac{\sqrt{n}\sigma _{\max }}{2\sqrt{m}},~c_4^*=c_5\frac{\sqrt{n}\sigma _{\max }^3}{2m^{3/2}}+c_6\frac{\sigma _{\max }^2}{m}. \end{aligned}$$

$\square$